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The Pell equation for $D = 14$

Example

The complete quotients of $\sqrt{14}$ are

\frac{\sqrt{14} + 3}{5}, \frac{\sqrt{14} + 2}{2}, \frac{\sqrt{14} + 2}{5}, \sqrt{14} + 3,

so $\sqrt{14} = [3; \overline{1, 2, 1, 6}] .$ The convergent $15 / 4$ is the first one with norm $1$ , so the fundamental Pell solution is $15 + 4 \sqrt{14} .$

Facts & Assumptions

Given: The positive nonsquare integer $D = 14$ .

[F1]

The continued fraction of $\sqrt{D}$ has symmetric period ending in $2 a_{0}$ , and the returned state satisfies $Q_{ℓ} = 1$ (The continued fraction of $\sqrt{D}$ has symmetric period ending in $2 a_{0}$ ).

[F2]

The convergents of $\sqrt{D}$ satisfy $p_{n}^{2} - D q_{n}^{2} = (- 1)^{n + 1} Q_{n + 1} .$ (Convergents to $\sqrt{D}$ satisfy the norm identity).

Verification

technique · direct

1.1

Starting from $\sqrt{14}$ , rationalizing reciprocals gives the state cycle $α_{1} = \frac{\sqrt{14} + 3}{5}, α_{2} = \frac{\sqrt{14} + 2}{2}, α_{3} = \frac{\sqrt{14} + 2}{5}, α_{4} = \sqrt{14} + 3,$ so the digits are $1, 2, 1, 6$ and $\sqrt{14} = [3; \overline{1, 2, 1, 6}] .$ The convergents are therefore $3, 4, \frac{11}{3}, \frac{15}{4} .$

givenalgebra

2.1

Here $ℓ = 4$ and $Q_{4} = 1$ , in agreement with [F1]. Applying [F2] at $n = 3$ gives $15^{2} - 14 \cdot 4^{2} = (- 1)^{4} Q_{4} = 1,$ so $15 + 4 \sqrt{14}$ is the least positive norm-one solution.

F1F2step 1.1algebra∎

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Sources

Peter Hackman, Elementary Number Theory (standard reference, not scraped)
MIT 18.781, Lecture 21: Brahmagupta-Pell Equation (standard reference, not scraped)

The Pell equation for D=14D=14